Задача №1451

\[ \int\frac{\sqrt{1-x^2}}{x^2}dx =\left[\begin{aligned}& x=\sin{t}.\\& dx=\cos{t}dt.\end{aligned}\right] =\int\frac{\cos{t}\cdot\cos{t}dt}{\sin^2{t}} =\int\frac{\cos^2{t}dt}{\sin^2{t}} =\int\frac{1-\sin^2{t}}{\sin^2{t}}dt =\int\left(\frac{1}{\sin^2{t}}-1\right)dt=\\ =-\ctg{t}-t+C =-\frac{\cos{t}}{\sin{t}}-t+C =-\frac{\sqrt{1-\sin^2{t}}}{\sin{t}}-t+C =-\frac{\sqrt{1-x^2}}{x}-\arcsin{x}+C \]